Find q when 22.0 g of water is heated from 45.0°C to 100.0°C.?
- jacob sLv 74 weeks agoFavorite Answer
We know the relationship
q = m*s*∆T
Where q is heat, m is mass of substance, s is specific heat
∆T is change in temperature
m = 22.0 g
specific heat of water, s = 4.184 Jg-1oC-1
∆T = (100-45)oC = 55oC
q = 22.0 g *4.184 Jg-1oC-1 *55oC
q = 5062.64 J
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